Strategy in solving vector equations involving grad, scalar product operators?

[jason12345]

What is the general strategy in solving vector equations involving grad and the scalar product?

In particular, I want to express [itex]\Lambda[/itex] in terms from [itex] \mathbf U \cdot \nabla\Lambda = \Phi[/itex] but it looks impossible, unless there is some vector identity I can use.

Thanks in advance.

 

[I like Serena]

Hi jason12345!

What you have there is a directional derivative.
See for instance wiki: http://en.wikipedia.org/wiki/Directional_derivative

If U is for instance a unit vector, then:
[tex]\vec U \cdot \vec\nabla \Lambda(\vec x) = \Phi(\vec x)[/tex]
is also written as:
[tex]\vec\nabla_{\vec U} \Lambda(\vec x) = \Phi(\vec x)[/tex]

You can find [itex]\Lambda[/itex] with for instance something like:
[tex]\Lambda(\vec x) = \int \Phi(\vec x + u \vec U) du[/tex]

 

[jason12345]

Hi jason12345!

What you have there is a directional derivative.
See for instance wiki: http://en.wikipedia.org/wiki/Directional_derivative

If U is for instance a unit vector, then:
[tex]\vec U \cdot \vec\nabla \Lambda(\vec x) = \Phi(\vec x)[/tex]
is also written as:
[tex]\vec\nabla_{\vec U} \Lambda(\vec x) = \Phi(\vec x)[/tex]

You can find [itex]\Lambda[/itex] with for instance something like:
[tex]\Lambda(\vec x) = \int \Phi(\vec x + u \vec U) du[/tex]

That's given me a lot to think about - thanks!

Maybe I could try for something simpler to start with so:

How would I find [itex]\nabla\Lambda[/itex] as the other terms with [itex]U[/itex] independent of (x,y,z)?

Perhaps I could use some vector identity?

 

[ideasrule]

I don't think it's possible to solve. There's an infinite number of solutions, even after we disregard the ones that are off by a constant.

To illustrate this, suppose we want to construct a Λ such that U⋅∇Λ=Φ is satisfied. The left hand side is a sum of three terms: Ux*dΛ/dx, Uy*dΛ/dy, and Uz*dΛ/dz. For any point, we literally have absolute power to set dΛ/dx to whatever we want. After that, we have absolute power to choose any value for dΛ/dy. There's an infinite number of choices for dΛ/dx, and for each of those choices, there's an infinite number of choices for the other two derivatives.

This is analogous to a normal dot product, say a*(0,0,1)=0.5. The only thing this equation tells you is that the x component of a is 0.5. The y and z components can be anything, so it's not possible to solve for a in the sense of writing a concise analytical expression that captures all solutions.

 

[I like Serena]

In the example a*(1,0,0)=0.5 you can indeed say that the x-component is 0.5.
So the solution is a=(0.5,y,z) with 2 unknown variables y and z.


Similarly, if you know for instance that U=(1,0,0), the expression in the OP ([itex]\boldsymbol U \cdot \nabla \Lambda = \Phi[/itex]) reduces to:
[tex]\frac {\partial \Lambda} {\partial x} = \Phi[/tex]
So:
[tex]\Lambda = \int \Phi dx + C(y,z)[/tex]
And:
[tex]\nabla \Lambda = \begin{pmatrix} \Phi(x,y,z) \\ \frac {\partial} {\partial y} C(y,z) \\ \frac {\partial} {\partial z} C(y,z) \end{pmatrix}[/tex]

You can also see here that it is not simpler to solve for [itex]\nabla \Lambda[/itex].
You need to find [itex]\Lambda[/itex] first.


When U is not aligned with an axis, you will need to make a coordinate transformation in such a way that U is mapped to for instance (1,0,0).
Then you can use this same procedure.

As yet, I don't have a nice complete formula that includes the coordinate transformation though. That will take a bit of mind-benchpressing first.